# Opencv: Computing fundamental matrix from R and T

I want to compute the epipolar lines of a stereo camera. I know both camera intrinsics matrix as well as R and T. I tried to compute the essential matrix as told in Learning Opencv book and wikipedia.

where [t]x is the matrix representation of the cross product with t.

so

I tried to implement this with python and then use the opencv function cv2.computeCorrespondEpilines to compute the epilines.

The problem is that the lines I get don't converge in a point as they should... I guess I must have a problem computing F.

This is the relevant pice of code:

T #Contains translation vector R #Rotation matrix S=np.mat([[0,-T[2],T[1]],[T[2],0,-T[1]],[-T[1],T[0],0]]) E=np.mat(R)*S M1=np.mat(self.getCameraMatrix(cam1)) M1_inv=np.linalg.inv(M1) M2=np.mat(self.getCameraMatrix(cam2)) M2_inv=np.linalg.inv(M2) F=(M2_inv.T)*E*M1_inv

The matrices are:

M1=[[ 776.21275864 0. 773.70733324] [ 0. 776.21275864 627.82872456] [ 0. 0. 1. ]] M2=[[ 764.35675708 0. 831.26052677] [ 0. 764.35675708 611.85363745] [ 0. 0. 1. ]] R=[[ 0.9999902 0.00322032 0.00303674] [-0.00387935 0.30727176 0.9516139 ] [ 0.0021314 -0.95161636 0.30728124]] T=[ 0.0001648 0.04149158 -0.02854541]

The ouput F I get it's something like:

F=[[ 4.75910592e-07 6.28777619e-08 -2.78886982e-04] [ -4.66942275e-08 -7.62837993e-08 -7.34825205e-04] [ -8.86965149e-04 -6.86717269e-04 1.40633035e+00]]

EDITED: The cross multiplication matrix was wrong, it has to be: S=np.mat([[0,-T2,T1],[T2,0,-T[0]],[-T1,T[0],0]])

The epilines converge now at the epipole.

## Answers

Hum, your F matrix seems wrong - to begin with, the rank is closer to 3 than 2. From your data I get:

octave:9> tx = [ 0 -T(3) T(2) > T(3) 0 -T(1) > -T(2) T(1) 0] tx = 0.000000 0.028545 0.041492 -0.028545 0.000000 -0.000165 -0.041492 0.000165 0.000000 octave:11> E= R* tx E = -2.1792e-04 2.8546e-02 4.1491e-02 -4.8255e-02 4.6088e-05 -2.1160e-04 1.4415e-02 1.1148e-04 2.4526e-04 octave:12> F=inv(M1')*E*inv(M2) F = -3.6731e-10 4.8113e-08 2.4320e-05 -8.1333e-08 7.7681e-11 6.7289e-05 7.0206e-05 -3.7128e-05 -7.6583e-02 octave:14> rank(F) ans = 2

Which seems to make more sense. Can you try that F matrix in your plotting code?